3

$G=(GL_2(\mathbb R),.)$ be group of all invertible $2x2$ matrices with coefficients in $\mathbb R$ under multiplication.

Find the $Z(G)$

I am having trouble with finding it. We know that elements of $GL_2(\mathbb R)$ have non-zero determinant. Namely, $A\in GL_2(\mathbb R)$ such that $A=\begin{pmatrix} a &b\\ c & d \end{pmatrix}$ then $ad-bc\not=0$. And if $\begin{pmatrix} x & y\\ z & t \end{pmatrix}$ in the center of $G$

Then:

$\begin{pmatrix} x & y\\ z & t \end{pmatrix}\begin{pmatrix} a &b\\ c & d \end{pmatrix}=\begin{pmatrix} xa+yc & xb+yd\\ za+tc & zb+td \end{pmatrix}=\begin{pmatrix} ax+bz & ay+bt\\ cx+dz & cy+dt \end{pmatrix}=\begin{pmatrix} a &b\\ c & d \end{pmatrix}\begin{pmatrix} x & y\\ z & t \end{pmatrix}$

It follows that:

  • $bz=yc$
  • $xb+yd=ay+bt$
  • $cx+dz=za+tc$

As you can see it is very messy, is there any method avaible that help us to find the center (method can be advanced or elementary.).

Edit:

I didnot know this question has duplicate (sorry about it). However, in given links these are prove-disprove question, I rather than wonder that why someone thinks that center of $GL_2(R)$ should be $Z\bigl(GL(2,\mathbb R)\bigr)= \left\{ \pmatrix{ a & 0 \\ 0 & a } \,\middle|\, a \in \mathbb R\setminus\{0\} \right\}$. Where did the thought come from? Is it too stupid to be curious about ? :)

Micheal Brain Hurts
  • 2,198
  • Possible duplicate of Proof of center of $GL(2,\mathbb R)$. – Dietrich Burde Jun 09 '18 at 16:26
  • Thank you but it is already given, just wanted someone to prove it. How and where does it come from? Why one thinks that $Z\bigl(GL(2,\mathbb R)\bigr)= \left{ \pmatrix{ a & 0 \ 0 & a } ,\middle|, a \in \mathbb R\setminus{0} \right}$. I am updating my question. – Micheal Brain Hurts Jun 09 '18 at 16:29
  • 1
    @userqrqh123: Your strategy is fine. But be aware that the element from the center is fixed, i.e. the values of $x,y,z,t$ are fixed, and the equations have to be satisfied for any other matrix. In particular $$ \begin{pmatrix}1&1\0&1\end{pmatrix}.$$ – Jo Mo Jun 09 '18 at 16:31
  • 2
    (the matrix I wrote down then implies $z=0$ and $x=t$, and similarly you see $y=0$. Can you see this?) – Jo Mo Jun 09 '18 at 16:39
  • 1
    Yes, since your matrix is invertible, any center's element should commute with it, therefore the result comes very quickly. Thank you. – Micheal Brain Hurts Jun 09 '18 at 16:42

0 Answers0